1
$\begingroup$

Let $\mathbf{n}$ is a Gaussian random vector with mean $\mathbf{0}$ and co-variance matrix $\mathbf{H}$. Let $\mathbf{r} = Sign(\mathbf{n})$, where $Sign(n_i) = 1$ if $n_i>0$ and $Sign(n_i) = -1$ if $n_i<0$. Let $\mathbf{A}$ be a positive semi definite matrix. Let $\sigma = \mathbf{r}^{T}\mathbf{A}\mathbf{r}$. I wanted to know if it is possible to get a concentration inequality for $\sigma$, like

$P(\sigma>E\{\mathbf{n}^{T}\mathbf{A}\mathbf{n}\})\leq$ some function, f(N) of the dimensions of vector $\mathbf{r}$.

Any references also would be very helpful. Thanks in advance!

We can note that $E\{\mathbf{n}^{T}\mathbf{A}\mathbf{n}\} = trace (\mathbf{AH})$, $E\{\mathbf{r}^{T}\mathbf{A}\mathbf{r}\} =\frac{2}{\pi}trace (\mathbf{A}\tilde{\mathbf{H}})$, where $\tilde{\mathbf{H}} = arcSin(\mathbf{H})$, where $arcSin$ (inverse sine) is taken over each entry of $\mathbf{H}$ and $trace(\mathbf{AH})\leq trace(\mathbf{A}\tilde{\mathbf{H}})$ and entries of $\mathbf{H}$ are in the interval $[-1,1]$.

$\endgroup$
11
  • $\begingroup$ We, you will have Markov's inequality. Won't that be enough for whatever it is you want to do ? $\endgroup$ Jul 30, 2015 at 8:54
  • $\begingroup$ @Guillaume Dehaene Using the Markov inequality will give us the trivial bound $P(\sigma>E{\mathbf{n}^{T}\mathbf{A}\mathbf{n})\leq 1$, since $\frac{E{\mathbf{r}^{T}\mathbf{A}\mathbf{r}}{E{\mathbf{n}^{T}\mathbf{A}\mathbf{n}}\geq 1$ $\endgroup$
    – Rakshith
    Jul 30, 2015 at 8:58
  • $\begingroup$ Sorry, I am new to this. So having problems Latexing. $\endgroup$
    – Rakshith
    Jul 30, 2015 at 9:06
  • $\begingroup$ I see where my mistake is. You're right that markov's bound is trivial in the case you are interested in. That does indicate that any bound you might find will be quite close to 1/2. Is that really an interesting result ? $\endgroup$ Jul 30, 2015 at 12:02
  • $\begingroup$ @Guillaume Dehaene Yes. Can you please give me a proof or reference. $\endgroup$
    – Rakshith
    Jul 30, 2015 at 12:48

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.